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Shkiper50 [21]
2 years ago
15

What is the symbol for decimeter

Chemistry
2 answers:
Marysya12 [62]2 years ago
5 0

Answer:

The symbol for decimeter is dm

Explanation:

No need for an explanation

alexgriva [62]2 years ago
3 0

Answer:

the symbol of decimeter is dm

Explanation:

hope this  helped have a good day :)

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Study the diagram about the varying pressures of Earth’s Layers.
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A) Outer core

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Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfr
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Answer:

a. 4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas N_2, oxygen gas O_2, water vapor H_2O and carbon dioxide CO_2. Let's write the decomposition of nitroglycerin into these 4 components:

C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by \frac{3}{2}:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by \frac{5}{2}:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

\frac{5}{2} + 6 = 8.5

This leaves 9 - 8.5 = 0.5 = \frac{1}{2} of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)

To make it look neater without fractional coefficients, multiply both sides by 4:

4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

V_{CO_2} = 41.0 L

T = -14.0^oC + 273.15 K = 259.15 K

p = 1 atm

Firstly, we may find moles of carbon dioxide produced using the ideal gas law pV = nRT.

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol

According to the stoichiometry of the balanced chemical equation:

4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g

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A.

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